find shifted coordinate in skewed square - image

Question

I have square, I know the X,Y coordinate for the (A,B,C,D) each, coordinate for (E,F,G,H) and the position for the circle inside first box (I,J).
so ..
I want to find the coordinates for the same circle inside the second box .. base on all the data have.

Answer 1

You need to find the transform from the first box to the second
B=T*A
so you need to find T which is a 3x3 matrix if this is on the plane
solve the equations as shown on this page http://andrew.gibiansky.com/blog/image-processing/image-morphing/
and he has the program too - you only need three points from the first quadrangle and the corresponding three points in the second quadrangle
private static float[] calculateTransform(Polygon pOriginal, Polygon pFinal){
float a = pFinal.xpoints[0];
float b = pFinal.ypoints[0];
float c = pFinal.xpoints[1];
float d = pFinal.ypoints[1];
float e = pFinal.xpoints[2];
float f = pFinal.ypoints[2];
float A = pOriginal.xpoints[0];
float B = pOriginal.ypoints[0];
float C = pOriginal.xpoints[1];
float D = pOriginal.ypoints[1];
float E = pOriginal.xpoints[2];
float F = pOriginal.ypoints[2];
float x = ((B-D)*(e-c) - (a-c)*(F-D)) / ((B-D)*(E-C) - (A-C)*(F-D));
float y = (a*(E-C) + A*(c-e) - c*E + e*C)/(A*(D-F) + B*(E-C) + C*F - D*E);
float t = c - x*C - y*D;
float z = ((B-D)*(f-d) - (b-d)*(F-D)) / ((B-D)*(E-C) - (A-C)*(F-D));
float w = (b*(E-C) + A*(d-f) - d*E + f*C)/(A*(D-F) + B*(E-C) + C*F - D*E);
float s = d - z*C - w*D;
float[] transform = {x, y, z, w, t, s};
return transform;
}
then apply T to any point on A to get the corresponding point on B
private static float[] applyTransform(float x, float y, float[] transform){
float a = transform[0];
float b = transform[1];
float c = transform[2];
float d = transform[3];
float t = transform[4];
float s = transform[5];
float p = a * x + b * y + t;
float q = c * x + d * y + s;
float[] result = {p, q};
return result;
}